Aja's favorite cereal is running a promotion that says $1$ -in- $4$ boxes of the cereal contain a prize. Suppose that Aja is going to buy $5$ boxes of this cereal, and let $X$ represent the number of prizes she wins in these boxes. Assume that these boxes represent a random sample, and assume that prizes are independent between boxes. What is the probability that she wins at most $1$ prize in the $5$ boxes? You may round your answer to the nearest hundredth. $P(X\leq1)=$
Explanation: Strategy (without a fancy calculator) The probability that Aja wins at most $1$ prize in the $5$ boxes is equivalent to the probability that she wins $0$ or $1$ prize. So we can find those probabilities and add them them together to get our answer: $\begin{aligned} P(X\leq1)&=P(\text{wins 0})+P(\text{wins }1) \\\\ &=P(X=0)+P(X=1) \end{aligned}$ Finding $P(X=0)$ Since $1$ -in- $4$ boxes contain a prize, we know $P({\text{win}})={25\%}$ and $P({\text{not}})={75\%}$. Winning $0$ prizes is equivalent to not winning in all $5$ boxes. We can multiply probabilities since we are assuming independence: $\begin{aligned} P(X=0)&=({0.75})({0.75})({0.75})({0.75})({0.75}) \\\\ &=({0.75})^5 \\\\ &\approx0.2373 \end{aligned}$ We'll come back and use this result later. Next, we need to find $P(X=1)$ (the probability that she wins $1$ prize). Finding $P(X=1)$ Winning $1$ prize in $5$ boxes means Aja needs to win a prize in $1$ box and not win a prize in $4$ boxes. Since $1$ -in- $4$ boxes contain a prize, we know $P({\text{win}})={25\%}$ and $P({\text{not}})={75\%}$. Since we are assuming independence, let's multiply probabilities to find the probability of winning $1$ prize followed by not winning $4$ prizes: $P({\text{W}}{\text{NNNN}})=({0.25})({0.75})^4\approx0.0791$ This isn't the entire probability though, because there are other ways to win $1$ prize from $5$ boxes (for example, NNNNW). How many different ways are there? We can use the combination formula: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _5\text{C}_1&=\dfrac{5!}{(5-1)!\cdot1!} \\\\ &=\dfrac{5 \cdot \cancel{4 \cdot 3 \cdot 2 \cdot 1}}{(\cancel{4 \cdot 3 \cdot 2 \cdot 1}) \cdot 1} \\\\ &=5 \end{aligned}$ There are $5$ ways to win $1$ prize from $5$ boxes. Do they all have the same probability? Each of the $5$ ways has the same probability that we already found: $\begin{aligned} P({\text{W}}{\text{NNNN}})&=({0.25})({0.75})^4\approx0.0791 \\\\ P({\text{N}}{\text{W}}{\text{NNN}})&=({0.25})({0.75})^4\approx0.0791 \\\\ P({\text{NN}}{\text{W}}{\text{NN}})&=({0.25})({0.75})^4\approx0.0791 \\\\ P({\text{NNN}}{\text{W}}{\text{N}})&=({0.25})({0.75})^4\approx0.0791 \\\\ P({\text{NNNN}}{\text{W}})&=({0.25})({0.75})^4\approx0.0791 \end{aligned}$ So we can multiply this probability by $5$ since that is how many ways there are to win $1$ prize from $5$ boxes: $\begin{aligned} P(X=1)&=5(0.25)(0.75)^4 \\\\ &\approx5(0.0791) \\\\ &\approx0.3955 \end{aligned}$ Putting it all together Let's return to our original strategy to answer the question: $\begin{aligned} P(X\leq1)&=P(\text{wins 0})+P(\text{wins }1) \\\\ &=P(X=0)+P(X=1) \\\\ &=(0.75)^5+5(0.25)(0.75)^4 \\\\ &\approx0.2373+0.3955 \\\\ &\approx0.6328 \end{aligned}$ The answer $P(X\leq1)\approx0.6328\approx0.63$